# consider the following subsets of complex plane

$$\Omega_1=\left\{c\in\Bbb C:\begin{bmatrix}1&c\\\bar c&1\\ \end{bmatrix}\text{ is non-negative definite } \right\}$$

$$\Omega_2= \left\{c\in\Bbb C: \begin{bmatrix} 1 & c & c \\ \bar c & 1 & c \\ \bar c & \bar c & 1 \\ \end{bmatrix} \text{ is non-negative definite } \right\}$$

Let $$\bar D=\{z\in\Bbb C:|z|\le1\}$$ Then 1) $\Omega_1=\bar D,\Omega_2=\bar D$

2). $\Omega_1\neq\bar D, \Omega_2=\bar D$

3). $\Omega_1=\bar D, \Omega_2\neq\bar D$

4). $\Omega_1\neq\bar D, \Omega_2\neq\bar D$

How to proceed? Thank you.

• First, recall the definition of a non-negative definite matrix. What conditions does this put on $c$ in each case? – Alex G. Jun 21 '15 at 17:48
• If these were real numbers, the first thing I would think of is that if the variance of each of two or three random variables is $1$, then the covariance is the correlation, and the correlation is always in $[-1,1]$, so $-1\le c\le 1$. Here one relies on the fact that the set of all matrices of covariances is precisely the set of all non-negative-definite matrices. My first guess is $|c|\le 1$ for both $\Omega_1$ and $\Omega_2$. ${}\qquad{}$ – Michael Hardy Jun 21 '15 at 17:53
• non negative means positive semi definite i.e. all eigen values are ≥0 and atleast one eigen value 0 – Chiranjeev_Kumar Jun 21 '15 at 17:54
• Just as a matter of style, I'd rather write $\{c\in\mathbb C:|c|\le 1\}$, using $c$ rather than $z$ as in the definitions of the other two sets. ${}\qquad{}$ – Michael Hardy Jun 21 '15 at 17:57
• @Michel Hardy : if we take $c=-1$ then 3by3 matrix is non singuler with determinant value -4 which contradicting the given condition – Chiranjeev_Kumar Jun 21 '15 at 18:24

1. Note that the matrices are Hermitian, so it is enough to check if the eigenvalues $\lambda\geq 0$ are non-negative, or equivalently, $$\mu~:=~1-\lambda~\leq~ 1.$$
2. The characteristic polynomials read $$p_1(\lambda) ~=~\mu^2-|c|^2,$$ and $$p_2(\lambda) ~=~\mu^3+|c|^2(2{\rm Re}(c) -3\mu),$$ respectively.
3. Define polar decomposition $c~=~re^{i\theta}~\in~\mathbb{C}$.
4. The roots are $$\mu~=~\pm |c|,$$ and $$\mu = 2 r \cos\frac{\theta+2\pi p}{3},\qquad p\in\mathbb{Z},$$ respectively.
5. Hence, $$\Omega_1~=~\{c\in \mathbb{C} \mid |c|\leq 1\}~=~\bar{D},$$ while $$\Omega_2~=~\{re^{i\theta}\in \mathbb{C} \mid \forall p\in \mathbb{Z}:~ 2 r \cos\frac{\theta+2\pi p}{3}\leq 1\}~\neq~\bar{D}.$$ It is straightforward to check that $$\{c\in \mathbb{C} \mid |c|\leq \frac{1}{2}\}~\subsetneq~\Omega_2~\subsetneq~\bar{D}.$$
The $2\times 2$ matrix has eigenvalues $1 \pm |c|$, so it is positive semidefinite if and only if $|c|\leq 1$.
For the $3\times 3$ matrix, observe that the determinant is $-4$ when $c=-1$, so the matrix is not positive semidefinite in this case.